Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 107 / 1

The weak maximum principle for elliptic operators says that if $c\le 0$ andthenIndeed, a positive interior maximum has $Du=0$ and $\Delta u\le 0$, so the differential inequality is incompatible with a strict positive maximum. Applying this argument to a standard strictly perturbed function and then letting the perturbation tend to zero handles equality and proves the weak statement.

SetThen $v\ge A$ and, because $1\le e^{x_1+d}\le e^{2d}$ and $c\le 0$,Thus $(\Delta +c)(u-v)=f-(\Delta +c)v\ge 0$ and $u-v\le 0$ on the boundary. The weak maximum principle for elliptic operators gives $u\le v$. Applying the same argument to $-u$ gives $|u|\le v$, hence

If $u_1,u_2$ are two solutions with the same boundary values, put $w=u_1-u_2$. The mean value theorem givesTherefore $\Delta w-qw=0$ with zero boundary data. Apply the weak maximum principle for elliptic operators to $w$ and $-w$ with zeroth-order coefficient $-q\le 0$. It follows that $w=0$, proving uniqueness.

The required estimate follows from one-dimensional barriers on a sufficiently narrow slab. Put $\kappa =\sqrt{\tau }$. On $|x_1|<d_1<\pi /(2\kappa )$, the functionsatisfies $h^{\prime \prime }+\tau h=0$, equals one at $x_1=\pm d_1$, and obeysChoose $\delta (\epsilon ,\tau )$ so thatand so that the analogous solution of $k^{\prime \prime }+\tau k=-1$ with zero endpoint values is bounded by $e^{2d_1}-1$. Comparing $u$ and $-u$ withand using $c\le \tau$ givesThe permitted width is of order ${\tau }^{-1/2}$, so it can be chosen with $\delta (\epsilon ,\tau )\to \infty$ as $\tau \downarrow 0$.